Question

Calculate the pH at 25degreeC of a 0.0053 M solution of a weak base with a K_b of 1.8 TIMES 10^-9 pH

Answer 1

Let α be the dissociation of the weak base
                            BOH <---> B ⁺ + OH^-

initial conc.            c               0         0

change               -cα            +cα      +cα

Equb. conc.         c(1-α)        cα      cα

Dissociation constant, K_b = (cα x cα) / ( c(1-α)               

                                          = c α² / (1-α)

In the case of weak bases α is very small so 1-α is taken as 1

So K_b = cα²

==> α = √ ( K_b / c )

Given K_b = 1.8x10^-9

          c = concentration = 0.0053 M

Plug the values we get α = 5.83x10^-4
So the concentration of [OH^-] = cα

                                           = 0.0053 x5.83x10^-4
                                           =3.09 x 10^-6 M

pOH = - log [OH^-]

        = - log  (3.09 x10^-6)

        = 5.5

So pH = 14 - pOH

           = 14 - 5.5

           = 8.5